我有一个非常困惑的数据，我注意到元素的末尾有 '\n' 的模式，在此之前它需要与单个元素合并。
示例列表:
ls = ['hello','world \n','my name','is john \n','How are you?','I am \n doing well']
ls
返回/试用:
print([s for s in ls if "\n" in s[-1]])
>>> ['world \n', 'is john \n'] # gave elements that ends with \n
如何让以 '\n' 结尾的元素与元素前的 1 合并？寻找这样的输出:
['hello world \n', 'my name is john \n', 'How are you?','I am \n doing well']
请您参考如下方法:
如果您正在减少一个列表，也许，一种可读的方法是使用reduce功能。
functools.reduce(func, iter, [initial_value]) cumulatively performs an operation on all the iterable’s elements and, therefore, can’t be applied to infinite iterables.
首先，你需要一种积累结果的方法，我使用了一个包含两个元素的元组:buffer with concatenated strings until I find "\n" 和列表结果。请参见初始结构 (1)。
ls = ['hello','world \n','my name','is john \n','How are you?','I am \n doing well']
def combine(x,y):
if y.endswith('\n'):
return ( "", x[1]+[x[0]+" "+y] ) #<-- buffer to list
else:
return ( x[0]+" "+y, x[1] ) #<-- on buffer
t=reduce( combine, ls, ("",[]) ) #<-- see initial struct (1)
t[1]+[t[0]] if t[0] else t[1] #<-- add buffer if not empty
结果:
['hello world \n', 'my name is john \n', 'How are you? ', 'I am \n doing well ']
(1) 解释了初始结构:你使用一个元组来存储缓冲区字符串直到 \n 和一个已经煮熟的字符串列表:
("",[])
意思是:
("__ buffer string not yet added to list __", [ __result list ___ ] )
